Question

The roots of the equation (b-c) x2 +2 (c-a)x +(a-b)=0 are always real and distinct

Answer 1

Solution:

Compare (b-c)x²+2(c-a)x+(a-b)=0 with Ax²+Bx+C=0, we

get

A = (b-c), B = 2(c-a), C=(a-b)

Discreminant (D) > 0

[ Since , roots are real and

distinct ]

B²-4AC > 0

=> [2(c-a)]² -4(b-c)(a-b)>0

=> 4(c²+a²-2ac)-4(ab-b²-ac+bc)>0

=> c²+a²-2ac-ab+b²+ac-bc>0

=> a²+b²+c²-ab-bc-ac>0

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